Prove that:$ \frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A $

To do:

We have to prove that \( \frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A \).

Solution:

We know that,

$\tan A=\frac{\sin A}{\cos A}$......(i)

$\cot A=\frac{\cos A}{\sin A}$......(ii)

Therefore,

$\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin A}{1-\frac{\cos A}{\sin A}}$

$=\frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\frac{\sin A}{\frac{\sin A-\cos A}{\sin A}}$

$=\frac{\cos ^{2} A}{\cos A-\sin A}+\frac{\sin ^{2} A}{\sin A-\cos A}$

$=\frac{\sin ^{2} A}{\sin A-\cos A}-\frac{\cos ^{2} A}{\sin A-\cos A}$

$=\frac{\sin ^{2} A-\cos ^{2} A}{\sin A-\cos A}$

$=\frac{(\sin A+\cos A)(\sin A-\cos A)}{\sin A-\cos A}$

$=\sin A+\cos A$

Hence proved.